Proof. We begin in the same way as in the proof of the Chain Rule: using Proposition 4.6, write \(f(x)-f(\ell ) = F_\ell (x) (x-\ell )\) for all \(x\) and in particular for \(x=g(y),\) so

\[ f(g(y)) - f(g(k)) = F_\ell (g(y)) (g(y) - g(k)). \]

Since \(f\) and \(g\) are inverse to each other, \(f(g(y))=y,\) so

\[ y-k = F_\ell (g(y))(g(y)-g(k)). \]

When \(y\ne k,\) we have \(y-k\ne 0,\) and the equation shows that \(F_\ell (g(y))\ne 0\) (because the left-hand side is not \(0\)). When \(y=k,\) we have \(F_\ell (g(y)) = F_\ell (\ell ) = f'(\ell )\) which is not \(0\) by assumption. Hence \(F_\ell (g(y))\) is never zero, and we can divide by it:

\[ \frac {1}{F_\ell (g(y))}(y-k) = g(y) - g(k). \]

Since \(\frac {1}{F_\ell (g(y))}\) is continuous at \(y=k\) by Algebra of Continuous Functions and continuity of composition, by Proposition 4.6 \(g(y)\) is differentiable at \(k\) with

\[ \frac {1}{F_\ell (g(k))} = \frac {1}{F_\ell (\ell )} = \frac 1{f'(\ell )} \]

as derivative. □

Proof. The function \(\ln (1+y)\) is differentiable (by Chain Rule), and its derivative at \(y=0\) is \(\frac 1{1+0}=1.\) By Proposition 4.6 we can write \(\ln (1+y) = F_0(y)y\) where the slope function \(F_0\) is continuous at \(0\) with \(F_0(0)=1.\) Then for a fixed \(x\in \RR ,\)

\[ \Bigl ( 1+\frac xn\Bigr )^n = e^{n \ln (1+\frac xn)} = e^{n F_0(\frac xn)\frac xn} = e^{F_0(\frac xn)x}. \]

The sequence \(y_n = \frac xn\) is monotone with limit \(0\) so Lemma 1.1 and its Corollary tell us that \(\lim _{n\to \infty } e^{F_0(y_n)x}\) is \(\lim _{y\to 0} e^{F_0(y)x}.\) Since \(e^{F_0(y)x}\) is a continuous function of \(y,\) by criterion of continuity its limit as \(y\to 0\) equals its value at \(0,\) which is \(e^{F_0(0)x} = e^{1\cdot x}=e^x.\) □

Proof. In Fig. 5.2, area(\(\triangle OP_0P_\alpha \)) \(=\) \(\frac 12\) \(\times \) base \(\times \) height \(=\frac 12\times 1 \times \sin \alpha = \frac 12 \sin \alpha .\)

The area of the sector \(OP_0P_\alpha \) with central angle \(\alpha \) is \(\frac \alpha {2\pi }\) \(\times \) area(circle) \(=\) \(\frac \alpha {2\pi }\times \pi =\frac 12\alpha .\)

Area(\(\triangle O P_0 T_\alpha \)) is \(\frac 12 \times 1 \times \tan \alpha .\) We have \(\triangle OP_0P_\alpha \) \(\subseteq \) sector \(OP_0P_\alpha \) \(\subseteq \) \(\triangle OP_0T_\alpha ,\) therefore \(\frac 12\sin \alpha \le \frac 12\alpha \le \frac 12\tan \alpha .\) □

Proof.

The inequality in Lemma 5.4 can be written as \(0\le |\sin x| \le x\) when \(x\in (0,\frac \pi 2),\) as \(\sin x\) is positive for these \(x.\) Since \(\lim _{x\to 0+} 0 = \lim _{x\to 0+}x = 0,\) we have \(\lim _{x\to 0+} |\sin x| =0\) by Sandwich Rule.

Putting \(t=-x,\) we have \(\lim _{x\to 0-} |\sin x| = \lim _{t\to 0+}|\sin (-t)| = \lim _{t\to 0+}|\sin t| = 0.\)

The one-sided limits exist and are equal, so \(\lim _{x\to 0}|\sin x|\) is their common value \(0.\) □

Proof of Lemma 5.6 — proof not examinable. If \(M\) is the midpoint of the segment \(P_\alpha P_\beta ,\) the vector \(\overrightarrow {OM}\) is \(\frac 12(\overrightarrow {OP}_\alpha +\overrightarrow {OP}_\beta )\) and so the \(y\)-coordinate of \(M\) is \(\frac 12(\sin \alpha +\sin \beta ).\) On the other hand, see Fig. 5.4, \(\overrightarrow {OM}\) is proportional to \(\overrightarrow {OP}_{(\alpha +\beta )/2}\) and, from the right-angled triangle \(\triangle OMP_\beta \) we can see that \(\overrightarrow {OM} = \cos (\frac {\alpha -\beta }2)\overrightarrow {OP}_{(\alpha +\beta )/2}.\) This expresses the \(y\)-coordinate of \(M\) via the \(y\)-coordinate of \(P_{(\alpha +\beta )/2}:\)

\[ \frac 12(\sin \alpha +\sin \beta ) = \cos (\tfrac {\alpha -\beta }2) \sin (\tfrac {\alpha +\beta }2). \]

This sine addition formula holds for all \(\alpha ,\beta \in \RR .\) Substitute \(\alpha =y,\) \(\beta =-x\) to obtain the sine subtraction formula i. as claimed.

Substitute \(\alpha =\frac \pi 2-y,\) \(\beta =x-\frac \pi 2:\) the LHS becomes \(\frac 12(\sin (\frac \pi 2-y)-\sin (\frac \pi 2-x))\) which is \(\frac 12(\cos y-\cos x),\) the RHS becomes \(\cos (\frac \pi 2-\frac {y+x}{2})\sin (\frac {-y+x}2)=-\sin (\frac {y+x}2)\sin (\frac {y-x}2),\) equivalent to the cosine subtraction formula ii. □

Proof. In the sine subtraction formula (Lemma 5.6), we bound the modulus of \(\cos \) by \(1:\)

\[ |\sin y - \sin x| = 2 \left |\sin \tfrac {y-x}2\right | \left |\cos \tfrac {y+x}2\right | \le 2 |\sin h|, \]

where we put \(h=(y-x)/2.\) Opening out the modulus, we get

\[ -2|\sin h| \le \sin y - \sin x \le 2|\sin h|. \]

Taking the limit as \(y\to x,\) equivalently \(h\to 0,\) we have \(\lim _{h\to 0} 2|\sin h|=0\) by Corollary 5.5. Hence by Sandwich Rule

\[ \lim _{y\to x} \sin (y)-\sin (x) = 0 \quad \Leftrightarrow \quad \lim _{y\to x} \sin (y) =\sin (x), \]

so by the criterion of continuity, \(\sin \) is continuous at \(x.\)

Continuity of \(\cos \) is proved similarly and is left to the student. □

Proof. We compute the one-sided limits. If \(x\in (0,\frac \pi 2),\) we have

\[ \cos x = \frac {\sin x}{\tan x} \le \frac {\sin x}{x} \le \frac {\sin x}{\sin x}=1 \]

from the sine-angle-tangent inequality in Lemma 5.4. As \(x\to 0+,\) \(\cos x\to \cos 0=1\) because \(\cos \) is a continuous function. Hence by Sandwich Rule \(\lim _{x\to 0+}\frac {\sin x}x =1.\)

In \(\lim _{x\to 0-}\frac {\sin x} x,\) change the variable, \(x=-y.\) The limit becomes \(\lim _{y\to 0+}\frac {\sin (-y)}{-y}=\lim _{y\to 0+}\frac {\sin y} y=1.\)

Both one-sided limits are equal to \(1,\) so \(\lim _{x\to 0}\frac {\sin x}{x}\) exists and equals \(1,\) as claimed. □

Proof. Use the sine subtraction formula to write \(\sin y-\sin x =2\sin h \cos (x+ h)\) with \(h=(y-x)/2.\) Then

\[ \frac {\sin y-\sin x}{y-x} = \frac {2\sin h\cos (x+h)}{2h} = \frac {\sin h}{h} \cos (x+h), \]

and so

\[ \sin ' x = \lim _{h\to 0} \frac {\sin h}{h} \times \lim _{h\to 0} \cos (x+h) = 1\times \cos (x+0) =\cos x \]

by AoL, the Special Limit for sine and continuity of \(\cos .\)

To differentiate \(\cos ,\) use \(\cos y -\cos x = -2 \sin h \sin (x+ h)\) and conclude similarly. □